Question

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. $$3 x^{3}+7 y^{3}=10 y; \left(x_{0}, y_{0}\right)=(1,1)$$ (Graph cant copy)

Short answer

Question: Find the equations of the tangent and the normal line to the curve \(3x^3 + 7y^3 = 10y\) at the point \((1, 1)\) and describe the general appearance of the tangent and normal lines in relation to the curve. Answer: The equation of the tangent line is: $$y - 1 = \frac{9}{11}(x - 1)$$ The equation of the normal line is: $$y - 1 = -\frac{11}{9}(x - 1)$$. The tangent line touches the curve at point (1,1) with a somewhat steep but moderate positive slope of \(\frac{9}{11}\), while the normal line intersects the curve perpendicularly at the same point with a somewhat steep negative slope of \(-\frac{11}{9}\).

Step-by-step solution

Find the derivative dy/dx using implicit differentiation

We start by differentiating the given equation with respect to x using implicit differentiation. $$\frac{d}{dx} (3x^3 + 7y^3) = \frac{d}{dx}(10y)$$ Differentiate both sides: $$ 9x^2 + 21y^2\frac{dy}{dx} = 10\frac{dy}{dx} $$ Now, we'll solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}(21y^2 - 10) = 9x^2$$ $$\frac{dy}{dx} = \frac{9x^2}{21y^2 - 10}$$

Evaluate the derivative at the point (1, 1) to find the slope of the tangent line

We want to find the slope of the tangent line at the point \((x_0, y_0) = (1, 1)\). To do this, we plug the point into our derivative equation: $$\frac{dy}{dx} (1,1) = \frac{9(1)^2}{21(1)^2 - 10} = \frac{9}{11}$$ So, the slope of the tangent line at \((1,1)\) is \(\frac{9}{11}\).

Determine the equation of the tangent line using point-slope form

Now that we have the slope of the tangent line, we can use the point-slope form to find the equation of the tangent line. The point-slope form is: $$y - y_0 = m(x - x_0)$$ Plugging in our slope and point \((x_0, y_0) = (1, 1)\), we get: $$y - 1 = \frac{9}{11}(x - 1)$$ This is the equation of the tangent line.

Find the slope of the normal line

As the normal line is perpendicular to the tangent line, its slope will be the negative reciprocal of the tangent line's slope. So: $$m_\text{normal} = -\frac{11}{9}$$

Determine the equation of the normal line using point-slope form

Using the point-slope form with the normal slope found in Step 4, we get the equation of the normal line as: $$y - 1 = -\frac{11}{9}(x - 1)$$

Describe the graph

Since we can't copy or draw the graph, we can describe the general appearance of the tangent and normal lines in relation to the curve. The tangent line is touching the curve at point (1,1) with a slope of \(\frac{9}{11}\), indicating a somewhat steep but moderate positive slope. The normal line intersects the curve perpendicularly at the same point with a slope of \(-\frac{11}{9}\), indicating a somewhat steep negative slope. To summarize, the equation of the tangent line is: $$y - 1 = \frac{9}{11}(x - 1)$$ The equation of the normal line is: $$y - 1 = -\frac{11}{9}(x - 1)$$.

Key concepts

Implicit Differentiation

Implicit differentiation serves as a technique used when dealing with equations that define one variable, typically 'y', implicitly in terms of another variable, like 'x'. In calculus, when you encounter a complex equation where 'y' cannot be easily isolated, implicit differentiation allows you to find the derivative \(\frac{dy}{dx}\) without explicitly solving for 'y'.

Here's a step-by-step guide to implement this method:

• Differentiate both sides of the equation with respect to 'x', treating 'y' as a function of 'x'.
• Remember that when you differentiate 'y' with respect to 'x', you apply the chain rule, which adds \(\frac{dy}{dx}\) to the derivative.
• Rearrange the equation to solve for \(\frac{dy}{dx}\), which gives you the slope of the tangent line at any point on the curve.

Using implicit differentiation makes finding the slope at a specific point, like \( (x_0, y_0) = (1,1) \), straightforward. Simply substitute the 'x' and 'y' values into the derived formula.

Slope of Tangent Line

The slope of a tangent line to a curve at a given point is simply the value of the derivative of that curve at that point. The slope is a measure of how steep the line is at the point of tangency.

To find it, take these steps:

• Calculate \(\frac{dy}{dx}\) using the appropriate method, such as implicit differentiation if needed.
• Evaluate the derivative at the given point to ascertain the slope.
• If \(\frac{dy}{dx}\) at the point \( (x_0,y_0)\) is \(m\), then the slope of the tangent line at that point is 'm'.

In the exercise example, after using implicit differentiation, evaluating the derivative at \( (1,1)\) provides us with \(\frac{9}{11}\), the slope of the tangent line at that point.

Point-Slope Form

The point-slope form is a format used to write the equation of a line when you are given a point on the line \( (x_0, y_0)\) and the slope of the line 'm'. The formula is expressed as:
\[y - y_0 = m(x - x_0)\]
It's a pivotal tool in calculus for converting the slope of the line and a reference point into an equation.

For instance, using the point-slope form with \( (x_0,y_0) = (1,1)\) and the slope \(m=\frac{9}{11}\), as we found from the previous concept, we can express the line equation as: \[y - 1 = \frac{9}{11}(x - 1)\]. This equation represents the tangent line that just 'grazes' the curve at the point \( (1,1)\).

Slope of Normal Line

A normal line to a curve at a given point is a straight line that is perpendicular to the tangent line at that point. The slope of the normal line is the negative reciprocal of the slope of the tangent line, which can be expressed as \(m_{normal} = -\frac{1}{m_{tangent}}\).

For example, if the slope of the tangent line is \(\frac{9}{11}\), then the slope of the normal line at that point would be \(m_{normal} = -\frac{11}{9}\). With this slope, and again using the point-slope form with the same point \( (x_0, y_0) = (1,1)\), the equation of the normal line becomes: \[y - 1 = -\frac{11}{9}(x - 1)\].

The normal line and tangent line equations provide two perpendicular lines that intersect at the curve, giving a clear visual of how the curve behaves at the tangent point.